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Ans: D

As the position of the speakers matter, this is permutation.
So total available spaces: 6!
Total positions: 4!
Thus, 6!/4! = 30
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So we have 6 positions and a condition that 4 speakers should be in certain order in the event.

So imagine that we have 6 empty seats and now we have to pick 2 seats for the remaining 2 person who can sit anywhere ( because they are not constrained) so 6c2 = 15.

So now we 2 speakers can be arranged in 2! = 2 different ways

so 15 * 2 = 30

So D is the correct answer
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There are two methods of solving:
1. Consider two cases:
1. The two other speakers speak consecutively: They can either start, or speak in between different speakers, or end. Totally, there are 5 spaces that can be fit. Also, they can have an internal permutation of 2!. Therefore that comes to 5*2! = 10
2. The other two speakers do not speak consecutively: The 2 speakers can be arranged in 5 spaces, which comes to 5P2 = 20
Total = 10 + 20 = 30

2. The total ways that the speakers can be arranged is 6!. 4 of those speakers have a particular ordering so 1 out of 4! ways is permissible. Therefore total arrangements = 6!/4! = 720/24 = 30.

Answer D
Bunuel
A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?

A. 6
B. 12
C. 24
D. 30
E. 120

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ABCD sequence always has to be in this way. That means = 6!/4! = 30 (D)
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P(n,n)=n!=6!
6!=6x5x4x3x2x1=720
A before B before C before D
6!/4!=720/24=30
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So we have 6 spots and for 6 speakers
4 speaker arrangement is already provided , we just have to select 2 places from 6 spots for remaining 2 speaker

we can do it by 6C2 ways
but we have 2 remaining speaker so for each 2 spots we select we have 2 possible arrangments possible for that spot

so our answer is
2 *6c2
ANS Option D 30
Bunuel
A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?

A. 6
B. 12
C. 24
D. 30
E. 120

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Bunuel
A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?

A. 6
B. 12
C. 24
D. 30
E. 120

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six slots _ _ _ _ _ _
order which is already fixed A, B, C, D

Now choose 4 slots from 6 to place the fixed order. 6C4 which can be done in 15 ways
Now the remaining 2 slots are filled in 2 * 1 ways. = 2 ways

=> total ways 6 speakers can be scheduled with the given restriction is 30
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It helps to read between the lines here - Alex, Bruno, Carla, and Denise are 4 speakers with conditions among them (A goes before B, B before C, C before D). But there are 6 speakers; so, 2 of them aren't mentioned in the stem. Which means, there are no conditions defining them (except one speaker per slot, of course). I'll imagine them to be Eugene and Fred.

Now, ABCD must essentially be in this order and this order only, while EF can go anywhere.

To begin, let's find the number of ways we can arrange all 6 speakers - 6! - 6*5*4*3*2*1. That's 720.

Now, we can only really arrange ABCD in one possible order - or the order matters - or 6! (total ways to arrange) / 4! (total ways to pick four when the order matters) - 720 / 4*3*2*1 = 30.

That's our answer - D. 30.

Bunuel
A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?

A. 6
B. 12
C. 24
D. 30
E. 120

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Answer: D

Total no. of ways= 6!
Number of ways in which 6 speakers can be scheduled keeping in mind the conditions= \(6!/(2*3*4)\) = 30
Bunuel
A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?

A. 6
B. 12
C. 24
D. 30
E. 120

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There are 6 speakers , Total arrangement = 6! .
Alex , Bruno , Carla and Demise must keep their order fixed as
A<B<C<D
Among 4 people , Only 1 order is allowed out of 4! possible orders

So we divide by 4! possible order
6!/4!= 30 , Correct ans D .
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There are 6 possible ways to arrange all the speakers 6!. But the order of alex, bruno, carla and denise is fixed. Alex before bruno, bruno before carla and carla before denise so only 1 arrangement will be correct in this case so we must remove repetitions. Therefore 6!/4! = 30 Answer is D.
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Let's think of the schedule as 6 ordered slots: _ _ _ _ _ _

We must place Alex, Bruno, Carla and Denise in any of the 4 slots. Hence, we must first select 4 out of 6 slots.
Selecting 4 out of 6 slots = 6C4 = 6!/4!2! = 15

Now, we can arrange the remaining 2 speakers in any of the two remaining slots.
Arranging 2 items in 2 slots = 2! = 2

Therefore, total number of different ways the 6 speakers can be arranged = 15x2 = 30

Bunuel
A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?

A. 6
B. 12
C. 24
D. 30
E. 120

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  • Without any constraints, six speakers could be arranged in \( 6! = 720\) ways.

  • Within these 720 arrangements, without any constraints, four 'key' speakers A-B-C-D could be arranged in \(4! = 24\) ways.

However, since we know that among themselves, A-B-C-D only have ONE valid arrangement with \(A<B<C<D\), then we need to decrease the total number of permutations by all the 'extra' unconstrained and unsuitable permutations of the four key speakers:
\(X = 6! / 4! = 720/24 = 30\)

Therefore, the answer is D.
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It is given that there are 6 speakers total & among them, Alex, Bruno, Carla, and Denise, the order must be:

Alex comes first before Bruno followed by before Carla & then subsequently Denise

In all possible schedules of 6 speakers, the four can appear in any of relative orders, and only one of those orders is allowed.

So the number of valid schedules is:

6! / 4! = 720 / 24 = 30

Therefore the answer is option D
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There are 6 speakers including Alex, Bruno, Carla and Denise. Based on what is given, we know that in the order of speaking
Alex < Bruno < Carla < Denise

Now, total arrangements without any restriction for 6 speakers = 6! = 720
We also know that the 4 speakers above should appear in the exact relative order and not consecutively one after the other which is important to note here
So out of the 4! = 24 relative orders only one is allowed

We also know that, when say k distinct items appear in a fixed relative order
No of arrangements = n! / k!

Similarly,
Required number of arrangements = 6! / 4! = 30

D. 30
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As per the question, ABCD should have same sequence and rest two other speakers aren't known.

For arranging speakers in different ways as per the condition
We should treat ABCD as a single entity and then 2 other speakers individually

Now the total variables for us to arrange is 3 and can be done in 3! Ways.

Which is 6
Bunuel
A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?

A. 6
B. 12
C. 24
D. 30
E. 120

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Order of speaking we get

A -> B -> C -> D

Spacing them we get

_ A _ B _ C _ D _

We have five spaces, and we take the below scenarios

1) The remaining 2 people are together
Then we have 5 places to keep the remaining two people and also for interchanging their places
Total combinations = 5 * 2 = 10

2) The 1st person is kept fixed at 1st position and 2nd person goes through 2nd to 6th position
We get 5 positions and counting the reverse option we get 4 * 2 = 8

3) 1st person fixed at 2nd place and 2nd person goers through 3rd to 6th
We get 4 positions and counting the reverse option we get 3 * 2 = 6

Similarly for the other positions we get 4 and 2 combinations.

Total combinations = 10 + 8 + 6 + 4 + 2 = 30

Option D

Bunuel
A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?

A. 6
B. 12
C. 24
D. 30
E. 120

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